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Solve x(t) for 2x”(t)+6x’(t)+4x(t)=12, x(0)=3, x’(0)=1.
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The auxiliary equation is \(2m^2+6m+4=0 \implies m=-2\) or \(m=-1\). Thus the general solution is \(x_c=c_1e^{-2t}+c_2e^{-t}.\) Assume a particular solution \(x(t)=k\), where k is a constant. Then you can easily find that \(4k=12 \implies k=3\). Hence the general solution is \(x(t)=c_1e^{-2t}+c_2e^{-t}+3\). You can use the initial values given to find \(c_1\) and \(c_2\).
is that x(o)= 3 that i plug in? or the x'(o)= 1
Both of them.
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